Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{n^2 - 10n + 21}{5n - 35} \times \dfrac{n + 10}{-8n + 24} $
Solution: First factor the quadratic. $a = \dfrac{(n - 3)(n - 7)}{5n - 35} \times \dfrac{n + 10}{-8n + 24} $ Then factor out any other terms. $a = \dfrac{(n - 3)(n - 7)}{5(n - 7)} \times \dfrac{n + 10}{-8(n - 3)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (n - 3)(n - 7) \times (n + 10) } { 5(n - 7) \times -8(n - 3) } $ $a = \dfrac{ (n - 3)(n - 7)(n + 10)}{ -40(n - 7)(n - 3)} $ Notice that $(n - 7)$ and $(n - 3)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ \cancel{(n - 3)}(n - 7)(n + 10)}{ -40(n - 7)\cancel{(n - 3)}} $ We are dividing by $n - 3$ , so $n - 3 \neq 0$ Therefore, $n \neq 3$ $a = \dfrac{ \cancel{(n - 3)}\cancel{(n - 7)}(n + 10)}{ -40\cancel{(n - 7)}\cancel{(n - 3)}} $ We are dividing by $n - 7$ , so $n - 7 \neq 0$ Therefore, $n \neq 7$ $a = \dfrac{n + 10}{-40} $ $a = \dfrac{-(n + 10)}{40} ; \space n \neq 3 ; \space n \neq 7 $